title Sep 2021 revised Purpose Purpose of this exercise Lets consider how realistic the beam rifle in Gundam is based on the current accelerator technology by performing proton transport simulation in snowman ID: 1032253
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1. Melt a snowman by a proton beamtitleSep. 2021 revised
2. PurposePurpose of this exerciseLet’s consider how realistic the beam rifle in Gundam is based on the current accelerator technology by performing proton transport simulation in snowmanGeometry setupChange of sourceConcept of normalizationLecture onCourtesy of D. Satoh at JAEA
3. snowman.inpCheck Input FileBasic setupProjectile：Geometry：Tally：track.eps100-MeV proton (pencil beam with a radius of 1.0 cm)Water sphere of a 5-cm radius at the origin[t-track] fluence distribution[t-deposit] absorbed dose (Gy/source) in water spheredeposit.outDefinition of volume of sphere is necessary at [volume] section…x: Serial Num. of Regiony: Dose [Gy/source] p: xlin ylog afac(0.8) form(0.9)h: x n n y(all ),l3 n # num reg volume all r.err 1 1 5.2360E+02 2.9777E-11 0.002329.777 (pGy/source)
4. Sections in Input file[Title] Title of the simulation[Parameters] Define history number etc.[Source] Define source[Material] Define materials[Surface] Define surfaces[Cell] Define cells[Volume] Define volume of cells[T-Track] Draw particle trajectory[T-Deposit] Calculate deposition energysnowman.inpII. SourceI. GeometryIII. Tally (Detector)Sections in snowman.inp
5. Flow chart of this exerciseProcedureSet geometry of a snowmanSet beam conditionDetermine beam current and power to melt a snowman
6. Geometry setup of a snowmanInstructionA simple structure with one big and one small ice balls, and an aluminum plateMaterial of the ice balls is 1-g/cm3 water (without temperature option*)The aluminum plate is placed on top of the small ice ball\phits\utility\rotate3dshowGeometry setup in this exercise*Temperature option influence only motion of low-energy neutrons
7. Step 1: Construct a big ice ballSet a big ice ball (radius 20 cm, center at z = 0 cm) surrounding the original 10-cm sphereHintGeometry check can be done with icntl = 8A spherical surface centering the origin is defined by ”so radius”Define the region of the big ice ball not to overlap the region of the original sphere of 5-cm radius (avoid double defined region)Exclude a newly defined region (big ice ball) from voidCheck *_geo.out when you encounter an errorStep 1
8. Step 1[ P a r a m e t e r s] Icntl = 8 maxcas = 2000 maxbch = 1[ C e l l ] 1 1 -1.0 -1 $ Target 2 1 -1.0 -2 +1 $ Big ball98 0 #1 #2 -999 $ Void99 -1 999 $ Outer region[ S u r f a c e ] 1 so 5.0 2 so 20.0 999 so 200.0(1)(1) Geometry check (2) A spherical surface centering the origin(3) Define the region of the big ice ball (4) Exclude a newly defined region (2)(3)(4)Answer 1Set a big ice ball (radius 20 cm, center at z = 0 cm) surrounding the original 10-cm sphere
9. Step 2: Construct a small ice ballStep 2Set a small ice ball on top of the big ice ball (radius 15 cm, center at z = -25 cm)HintA spherical surface centering on z axis is defined by ”sz center-z-position radius” Exclude the region of big ice ball from small ice ball or vise versa, otherwise double defined region is created by two ice balls
10. Step 2[ C e l l ] 1 1 -1.0 -1 $ Target 2 1 -1.0 -2 +1 $ Big ball 3 1 -1.0 -3 +2 $ Small ball98 0 #1 #2 #3 -999 $ Void99 -1 999 $ Outer region[ S u r f a c e ] 1 so 5.0 2 so 20.0 3 sz -25.0 15.0 999 so 200.0(2)(1)(3)(1) A spherical surface centering on z axis(2) Define the region of the small ice ball (avoid double defined region)(3) Exclude the region of big ice ball from small ice ball Answer 2Set a small ice ball on top of the big ice ball (radius 15 cm, center at z = -25 cm)
11. Step 3: Set aluminum plateStep 3Define aluminum (Al) at [material] sectionPut an aluminum plate of a 10-cm radius and 4-cm thickness(-40 cm < z < -36 cm) on top of the small ballHintA cylindrical surface along the z axis is defined by “cz radius”A plane perpendicular to z axis is defined by ”pz z-position” Density of aluminum is 2.7 g/cm3 (negative value for mass density)Exclude region of the aluminum plate from that of the small ice ball
12. Step 3Define aluminum (Al) at [material] sectionPut an aluminum plate on top of the small ballAnswer 3[ M a t e r i a l ]MAT[ 1 ] # Water H 2.0 O 1.0MAT[ 2 ] # Al Al 1.0[ C e l l ] 1 1 -1.0 -1 $ Target 2 1 -1.0 -2 +1 $ Big ball 3 1 -1.0 -3 +2 +6 $ Small ball 4 2 -2.7 -4 +5 -6 $ Al plate98 0 #1 #2 #3 #4 -999 $ Void99 -1 999 $ Outer region[ S u r f a c e ] 1 so 5.0 2 so 20.0 3 sz -25.0 15.0 4 cz 10.0 5 pz -40.0 6 pz -36.0 999 so 200.0(1) Define aluminum at [material] section(2) A cylindrical surface along the z axis (3) A plane perpendicular to z axis (4) Define the region of aluminum plate (5) Exclude a newly defined region (2)(3)(1)(5)(4)
13. Step 4: Set proton beam conditionStep 4Find a proton beam energy so that the absorbed dose at central sphere is maximizedHintTransport calculation can be executed with icntl = 0Beam energy is given by e0 parameter at [source] sectionAbsorbed dose at central sphere can be checked by deposit.outProton absorbed dose is maximized at the Bragg peak
14. Step 4[ P a r a m e t e r s ]icntl = 0[ S o u r c e ]totfact = 1.0000 s-type = 1 proj = proton e0 = 200.00 r0 = 1.0000 (1)(2)Change the icntl parameterChange e0 parameterCheck the phits output files ・ track.eps ・ deposit.out (the 27th line)(3) Check the output file[ S o u r c e ]totfact = 1.0000 s-type = 1 proj = proton e0 = 293.00 r0 = 1.0000 Repeating these processes(2)Answer 4Find a proton beam energy so that the absorbed dose at central sphere is maximized
15. Default PHITS output is normalized to per particle emitted from sourceSet totfact at [source] section to change the normalization factor1 A = 1 C / 1 secondThe electric charge of a proton is 1.6x10-19CStep 5: Normalize the resultStep 5Calculate absorbed dose (Gy) by 1-second irradiation of a 10-nA proton beam, a conventional beam current for proton therapy HintMaxcas and maxbch control only statistical uncertainty, and they have no relation with normalizatione.g. Absorbed dose for 100 source particles is given by setting totfact = 100.0
16. Answer 5Answer to Step 5Check the scale change with totfactNumber of emitted protons for 1 ampere in 1 second is 1.0 / 1.6E-19 = 6.25E18 (particles)Number of emitted protons for 10 nA in 1 second is 6.25E18 x 10 x 1E-9 = 6.25E10 (particles)Absorbed dose with totfact = 6.25E10 is 1.283 (Gy) (for 293-MeV protons)
17. Step 6: Calculate beam current to melt central sphere by 1-second irradiationAssume the snowman is made of ice at -10CCalculate the proton beam current (A) and the power (MW) necessary to melt the central ice by 1 secondHint (assumption for simplicity)Specific heat of ice is 0.5 (cal/g/K) = 2.1 (J/g/K)Latent heat of ice (heat necessary for phase transition from ice to water) is 333.5（J/g）1Gy = 1（J/kg） = 0.001（J/g）*Beam power (MW) can be estimated by Particle energy (MeV) Beam current (A)For comparison…The maximum power of J-PARC (one of the most powerful accelerators in the world) is approximately 1 MW*In a strict sense, acceleration voltage (MV) x beam current (A)
18. Answer 6Answer to Step 6You can solve the problem by answering the following questions step by stepHow large is the absorbed dose (J/g) in central sphere by a 293-MeV proton beam with 10 nA in 1 second? How large is the heat (J/g) needed to heat up the ice by 10 K and melt the ice?What current (A) is required to give that heat in 1 second?How large is the power (MW) at this beam current?Melting a snowman is all we can do with current acceleratorsBeam rifle in Gundam is far beyond current technology !(Of course, with longer time, we can melt metals)About 1-MW beam power is required to melt a snowman(Details are given in answer\answer-snowman-en.ppt)
19. Construct geometry and tallyDefine source particlesNormalize the tally resultsSummarySummaryPHITS simulation should be conducted with following order: